To solve this problem, we can use the relationship between the force of friction and the normal force.

Given:

• Horizontal force $F = 100 \, \text{N}$
• Mass of the box $m = 50 \, \text{kg}$
• Gravitational acceleration $g = 9.8 \, \text{m/s}^2$ (standard value)

At constant speed, the force applied is equal to the frictional force. So, the frictional force $F_{\text{friction}} = 100 \, \text{N}$.

The formula for the frictional force is:

$F_{\text{friction}} = \mu \cdot F_{\text{normal}}$

where:

• $\mu$ is the coefficient of friction (what we need to find)
• $F_{\text{normal}}$ is the normal force, which is equal to the weight of the box in this case.

The normal force is given by:

$F_{\text{normal}} = m \cdot g = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 490 \, \text{N}$

Now, using the friction equation:

$100 \, \text{N} = \mu \cdot 490 \, \text{N}$

Solving for $\mu$:

$\mu = \frac{100}{490} \approx 0.204$

The coefficient of friction ($\mu$) is approximately 0.204.

Would you like more details on the steps, or do you have any further questions?

Here are 5 related questions to deepen your understanding:

1. What would happen to the coefficient of friction if the mass of the box increased?
2. How would the frictional force change if the box were pulled on a sloped surface?
3. What is the difference between static and kinetic friction, and which one is involved here?
4. How does the coefficient of friction depend on the materials in contact?
5. How would the problem change if the surface were lubricated?

Tip: When an object is moving at constant speed, the net force acting on it is zero. This means that any applied force exactly balances the opposing forces, like friction.